#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
template<class T> bool chmax(T &a, const T &b) { if(a<b){a=b;return 1;}return 0;}
template<class T> bool chmin(T &a, const T &b) { if(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

int T;
ll N,X;
ll dp[22][22][2][2][2];
ll p10[20];
const ll mo=998244353;
void solve() {
	int i,j,k,l,r,x,y; string s;
	
	p10[0]=1;
	FOR(i,19) p10[i+1]=p10[i]*10;
	
	cin>>T;
	while(T--) {
		cin>>N>>X;
		ZERO(dp);
		dp[0][0][1][1][0]=1;
		for(int d=0;d<=18;d++) {
			y=N/p10[d]%10;
			for(x=0;x<=18;x++) {
				int a,b,c,ad,bd;
				FOR(a,2) FOR(b,2) FOR(c,2) {
					FOR(ad,10) FOR(bd,10) {
						int na=a;
						int nb=b;
						if(ad>y) na=0;
						if(ad<y) na=1;
						if(bd>y) nb=0;
						if(bd<y) nb=1;
						(dp[d+1][x+(ad+bd+c>=10)][na][nb][ad+bd+c>=10]+=dp[d][x][a][b][c])%=mo;
					}
				}
			}
		}
		cout<<dp[19][X][1][1][0]<<endl;
	}
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}